Atcoder Grand Contest 046 D
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思路挺自然的一道 agc。
首先发现删除完字符后的状态可以用一个三元组 \\((i,j,k)\\) 表示，其中 \\(i\\) 表示删除完之后只剩 \\([i+1,n]\\) 的后缀，\\(j\\) 表示可以在后面插入 \\(j\\) 个 \\(0\\)，\\(k\\) 表示可以在后面插入 \\(k\\) 个 \\(1\\)，显然不同的三元组能够得到的串是不同的，而一组三元组可以得到的串的个数可以 DP 求解，具体来说，倒着 DP，同时为了防止算重，我们强制要求 \\(1\\) 的后面只能添加 \\(0\\)，\\(0\\) 的后面只能添加 \\(1\\)，这样转移是 \\(O(1)\\) 的，具体实现见代码。
接下来考虑哪些三元组 \\((i,j,k)\\) 可以得到。考虑正着 DP，那么发现每次有两种选择：
- 从 \\(s_i+1,s_i+2\\) 中保留一个，删除一个
- 从保留的字符中拿出一个 \\(s_i+1\\oplus 1\\)，插在开头，然后删除这个 \\(s_i+1\\oplus 1\\) 同时保留 \\(s_i+1\\)。
转移同样可以 \\(O(1)\\)，总复杂度 \\(O(n^3)\\)。
#include<bits/stdc++.h>
using namespace std;
const int MOD=998244353;
void add(int &x,int v)((x+=v)>=MOD)&&(x-=MOD);
int n,ok[305][305][305],dp[305][305][305],res;
char s[305];
int main()
scanf("%s",s+1);n=strlen(s+1);ok[0][0][0]=1;
for(int i=1;i<=n;i++)for(int j=n;~j;j--)for(int k=n;~k;k--)
ok[i][j][k]|=ok[i-1][j][k];
ok[i][j][k]|=ok[i][j+1][k];
ok[i][j][k]|=ok[i][j][k+1];
if(j&&i>=2&&(s[i]==\'0\'||s[i-1]==\'0\'))ok[i][j][k]|=ok[i-2][j-1][k];
if(k&&i>=2&&(s[i]==\'1\'||s[i-1]==\'1\'))ok[i][j][k]|=ok[i-2][j][k-1];
if(j&&s[i]==\'0\')ok[i][j][k]|=ok[i-1][j-1][k+1];
if(k&&s[i]==\'1\')ok[i][j][k]|=ok[i-1][j+1][k-1];
dp[n][0][0]=1;
for(int i=n;i;i--)for(int j=0;j<=n;j++)for(int k=0;k<=n;k++)
add(dp[i-1][j][k],dp[i][j][k]);
if(s[i]==\'0\')add(dp[i][j][k+1],dp[i][j][k]);
if(s[i]==\'1\')add(dp[i][j+1][k],dp[i][j][k]);
for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)for(int k=0;k<=n;k++)
if(ok[i][j][k])add(res,dp[i][j][k]);
add(res,MOD-1);printf("%d\\n",res);
return 0;
AtCoder Grand Contest 008 题解
A - Simple Calculator
模拟+分类讨论即可。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
if (ch == ‘-‘) ch = getchar(), a = -1;
else a = 1;
x = ch - ‘0‘;
while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
x = (x << 1) + (x << 3) + ch - ‘0‘;
return a * x;
}
int work(int x, int y)
{
if (x <= y) return y - x;
else return x - y + 2;
}
int main()
{
int x, y;
gii(x, y);
printf("%d
", min(work(x, y), min(1 + work(-x, y), min(1 + work(x, -y), 2 + work(-x, -y)))));
}
B - Contiguous Repainting
我们发现可以找到一段连续的k个，其它部分可以任意挑，连续的那一段要么全选要么全不选，枚举连续段在哪即可。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
if (ch == ‘-‘) ch = getchar(), a = -1;
else a = 1;
x = ch - ‘0‘;
while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
x = (x << 1) + (x << 3) + ch - ‘0‘;
return a * x;
}
int a[100010];
int64 s1[100010], s2[100010], ans;
int main()
{
int n, k;
gii(n, k);
for (int i = 1; i <= n; ++i) gi(a[i]), s1[i] = s1[i - 1] + a[i], s2[i] = s2[i - 1] + (a[i] > 0 ? a[i] : 0);
for (int i = 1; i <= n - k + 1; ++i)
{
int j = i + k - 1;
long long s = s2[i - 1] + s2[n] - s2[j];
s += max(s1[j] - s1[i - 1], 0LL);
ans = max(ans, s);
}
printf("%lld
", ans);
return 0;
}
C - Tetromino Tiling
我们发现只有1,2,4,5这四种才有可能组合出完整矩形，其中两个相同的可以组成4*2的，或者1 4 5各一个组成一个6*2的，两种情况取max就好。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
if (ch == ‘-‘) ch = getchar(), a = -1;
else a = 1;
x = ch - ‘0‘;
while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
x = (x << 1) + (x << 3) + ch - ‘0‘;
return a * x;
}
int a[7], b[7];
int64 ans = 0, res = 0;
int main()
{
for (int i = 0; i < 7; ++i) gi(a[i]), b[i] = a[i];
ans += 4LL * (a[0] >> 1); a[0] &= 1;
ans += 4LL * (a[3] >> 1); a[3] &= 1;
ans += 4LL * (a[4] >> 1); a[4] &= 1;
ans += 2LL * a[1]; a[1] = 0;
if (a[0] && a[3] && a[4]) ans += 6;
if (b[0] && b[3] && b[4]) --b[0], --b[3], --b[4], res += 6;
res += 4LL * (b[0] >> 1);
res += 4LL * (b[3] >> 1);
res += 4LL * (b[4] >> 1);
res += 2LL * b[1];
printf("%lld
", max(ans, res) >> 1);
}
D - K-th K
按坐标排序后贪心放就好了。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
if (ch == ‘-‘) ch = getchar(), a = -1;
else a = 1;
x = ch - ‘0‘;
while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
x = (x << 1) + (x << 3) + ch - ‘0‘;
return a * x;
}
PII x[510];
int a[510 * 510];
int n;
int main()
{
gi(n);
for (int i = 1; i <= n; ++i) gi(x[i].fi), x[i].se = i;
int k = 1;
sort(x + 1, x + n + 1);
for (int i = 1; i <= n; ++i)
{
a[x[i].fi] = x[i].se;
for (int j = 1; j < x[i].se; ++j)
{
while (a[k]) ++k;
a[k] = x[i].se;
}
if (k > x[i].fi)
{
puts("No");
return 0;
}
}
k = 1;
for (int i = 1; i <= n; ++i)
{
for (int j = x[i].se + 1; j <= n; ++j)
{
while (a[k]) ++k;
if (k < x[i].fi)
{
puts("No");
return 0;
}
a[k] = x[i].se;
}
}
puts("Yes");
for (int i = 1; i <= n * n; ++i)
printf("%d ", a[i]);
}
E - Next or Nextnext
不会做……先咕着。
F - Black Radius
计算出每个点和它父亲的d值最大多少是不同的，最后DP一下就好了。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
if (ch == ‘-‘) ch = getchar(), a = -1;
else a = 1;
x = ch - ‘0‘;
while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
x = (x << 1) + (x << 3) + ch - ‘0‘;
return a * x;
}
const int N = 2e5 + 10;
VI edge[N];
char str[N];
int n, in[N], out[N], dfs_clock, fa[N], dep[N], d[N];
void dfs1(int u)
{
in[u] = ++dfs_clock;
for (auto v : edge[u])
{
if (v == fa[u]) continue;
fa[v] = u;
dep[v] = dep[u] + 1;
dfs1(v);
}
out[u] = dfs_clock;
}
struct node
{
int l, r;
int val, tag;
void add(int x) { val += x; tag += x; }
} t[N << 2];
void push_up(int o) { t[o].val = max(t[o << 1].val, t[o << 1 | 1].val); }
void push_down(int o)
{
if (t[o].tag)
{
t[o << 1].add(t[o].tag);
t[o << 1 | 1].add(t[o].tag);
t[o].tag = 0;
}
}
void build(int o, int l, int r)
{
t[o].l = l, t[o].r = r;
if (l == r)
{
t[o].val = d[l];
return;
}
int mid = (l + r) >> 1;
build(o << 1, l, mid);
build(o << 1 | 1, mid + 1, r);
push_up(o);
}
int query(int o, int l, int r)
{
if (t[o].r < l || r < t[o].l) return 0;
if (l <= t[o].l && t[o].r <= r) return t[o].val;
push_down(o);
return max(query(o << 1, l, r), query(o << 1 | 1, l, r));
}
void add(int o, int l, int r, int x)
{
if (t[o].r < l || r < t[o].l) return;
if (l <= t[o].l && t[o].r <= r) return t[o].add(x);
push_down(o);
add(o << 1, l, r, x);
add(o << 1 | 1, l, r, x);
push_up(o);
}
/*
int pa[N];
int find(int x) { return x == pa[x] ? x : pa[x] = find(pa[x]); }
void unit(int x, int y)
{
x = find(x), y = find(y);
if (x == y) return;
pa[x] = y; str[y] |= str[x];
}
*/
int mxt[N], mxa[N], mxb[N];
void dfs2(int u)
{
int a = query(1, in[u], out[u]) + 1;
int b = max(query(1, 1, in[u] - 1), query(1, out[u] + 1, n));
if (fa[u]) mxt[u] = min(a + 1, b), mxa[u] = a + 1, mxb[u] = b; else mxt[u] = a; // >= t[u] : GG
for (auto v : edge[u])
{
if (v == fa[u]) continue;
add(1, in[v], out[v], -1);
add(1, 1, in[v] - 1, 1);
add(1, out[v] + 1, n, 1);
dfs2(v);
add(1, in[v], out[v], 1);
add(1, 1, in[v] - 1, -1);
add(1, out[v] + 1, n, -1);
}
}
int64 ans;
int num[N];
void dfs3(int u)
{
//cerr << u << ", " << mxt[u] << " : " << mxa[u] << ", " << mxb[u] << endl;
for (auto v : edge[u])
{
if (v == fa[u]) continue;
dfs3(v);
if (mxt[v] > num[v]) ans += mxt[v] - num[v], num[u] = min(num[u], min(mxa[v] - 1, mxb[v] + 1));
else num[u] = min(num[u], num[v]);
}
}
int main()
{
gi(n);
for (int i = 1; i < n; ++i)
{
int u, v;
gii(u, v);
edge[u].pb(v);
edge[v].pb(u);
}
scanf("%s", str + 1);
for (int i = 1; i <= n; ++i) str[i] -= ‘0‘;
for (int i = 1; i <= n; ++i) if (!str[i]) num[i] = N;
int root = 1;
for (; !str[root]; ++root);
dfs1(root);
for (int i = 1; i <= n; ++i) d[in[i]] = dep[i];
build(1, 1, n);
dfs2(root);
dfs3(root);
printf("%lld
", ans + mxt[root]);
return 0;
}
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